Lue Vang Mr. Kellogg AP Physics - Pd. 7 6 April 2012 Performed On: 4-5 April 2012 Lab Partners: Sir Sebastian Alvarez, Esq., Lt. Lue Vang
Torque and Rotational Equilibrium
Purpose: This lab will study torque, specifically balancing two levels of torque on two sides of a pivot point to achieve rotational equilibrium. The lab will study the effects of having variable masses at variable distances from the pivot point.
Background: The lever arm will be the distance of the placed mass away from the pivot point. The percent difference between two of the same variables of the same units is the difference between the two values divided by the average of the two values.
Hypothesis: Since rotation is caused by a net torque, balancing the torques on either side of the fulcrum should result in a stable, balanced meter stick. Since torque is r⊥F, moving the masses further from the fulcrum should increase the torque on that side of fulcrum. As it is, a light weight can be balanced with a large weight by placing the light weight closer to the edge of its side and the large weight closer to the balance point.
Apparatus:
Meter stick (1)
Unknown masses (4+)
Attachable hangers (4+)
Fulcrum mechanism (1)
Procedure: Case I: Two Known Masses: First, balance a meter stick on the fulcrum. Hang a 100g mass at 15.0cm on the meter stick and hang a 200g mass on the other side so that the meter stick is balanced on the fulcrum. Record the data and calculate torque. Case II: Three Known Masses: First, balance a meter stick on the fulcrum. Hang a 100g mass at 30.0cm on the meter stick. Then, hang 200g at 70.0cm. Finally hang a 50g mass in a position that will balance the meter stick on its fulcrum. Record the data and calculate torque. Case III: Unknown Masses: First, balance a meter stick on the fulcrum. Place one of the unknown masses at 10.0cm and then suspend a 200g mass on the other side so that the meter stick is balanced. Calculate a prediction for the position of the 200g and then test it. Measure and record data. Case IV: Meter Stick with One Mass: Suspend 100g at 0.0cm of the meter stick. Slide the fulcrum position on the meter stick so that the meter stick is balanced. Record the data and calculate toqrues.
Data: Case I: Two Known Masses:
Pivot Point (m):
0.502
Table 1.0: Two Known Masses
Mass #
Mass (kg)
Mass Position (m)
Lever Arm (m)
Weight Force (N)
Torque (m*N)
m1
0.097
0.150
0.352
0.951
0.335
m2
0.201
0.671
-0.169
1.970
-0.333
lever arm = mass position – pivot point = m i.e.) 0.502 – 0.150 = 0.352 m weight = mass * gravity = mg = kg*m/s2 = N i.e.) 0.097 * 9.8 = 0.951 N torque = lever arm * weight = rF = m*N i.e.) -0.352 * 0.951 = -0.335
rotational equilibrium = 0 = τccw + τcw … … -τcw = τccw -( -0.098*1.970) + ((0.502-x) * 0.490)= (0.302 * 0.951) x = .694 m *x = calculated mass position
Table 2.4: Percent Error
Calculated Mass Position (m)
Actual Mass Position (m)
Percent Error (%)
0.694
0.690
0.580
percent error = |(theoretical value – actual value)/actual value|*100 i.e.) |(0.694 – 0.690)/.690|*100 = 0.58%
Case III: Unknown Masses:
Pivot Point (m):
0.502
Moving Mass (kg):
0.201
Weight of Moving Mass (N):
1.970
Mass of Hanger (kg):
0.017
Table 3.0: Position of Stationary Masses
Stationary Mass #
Stationary Mass Position (m)
Lever Arm of Stationary Mass Position (m)
B
0.100
0.402
D
0.100
0.402
E
0.100
0.402
F
0.100
0.402
R
0.100
0.402
Table 3.1: Position of Moving Masses
Stationary Mass #
Moving Mass Position (m)
Lever Arm of Moving Mass (m)
B
0.969
-0.467
D
0.983
-0.481
E
0.946
-0.444
F
0.929
-0.427
R
0.801
-0.299
*see Table 1.0 for lever arm calculations *Mass B had .04kg added to be able to fit it onto the meter stick
Table 3.2: Calculate Mass
Stationary Mass #
Predicted Mass Value
Actual Mass Value
B
0.257
0.257
D
0.224
0.271
E
0.205
0.206
F
0.197
0.197
R
0.133
0.132
calculated stationary mass = actual mass + mass of hanger = ma + mh = ((lever arm of moving mass * weight of moving mass)/(lever arm of stationary mass*9.8)) - mass of hanger = ((rm * wm)/(rs * g)) - mh = (-(rm * 1.97)/(0.402 * .98)) – 0.017 i.e.) (-(-0.467 * 1.97)/(0.402 * .98)) – 0.017 = .216
Table 3.3: Percent Error
Stationary Mass #
Percent Error (%)
B
0.19
D
17.52
E
0.47
F
0.24
R
0.39
*see Table 2.4 for percent error calculations
Case IV: Meter Stick with One Mass:
Pivot Point (m):
0.293
Actual Mass of Meter Stick (kg):
0.135
Table 4.0: Meter Stick with One Mass Data
Mass #
Mass (kg)
Mass Position (m)
Lever Arm (m)
Weight Force (N)
Torque (m*N)
m1
0.097
0.000
0.293
0.951
0.279
m2
0.135
0.500
-0.207
1.323
-0.274
*see Table 1.0 for calculations
Table 4.1: Percent Difference
Counter Clockwise Torque (m*N)
Clockwise Torque (m*N)
Difference of Magnitudes (m*N)
Average (m*N)
Percent Difference (%)
0.279
-0.274
0.005
0.28
1.69
*see Table 2.1 for Percent Difference Calculations
Analysis: According to the data, the torque from a large force can be reduced by moving the force closer to the axis. In vice versa, a small force can produce a larger torque by applying that force farther away from the axis.
Note though, that the meter stick and hanging masses are suspended above the table; thus, there is an upward force equal to all the downward force from gravity in order for the objects to remain elevated. These “support forces” are not included in torque calculations because they cancel out and have a net force of zero, thus no net torque.
All in all, if an object is balanced at rotational equilibrium, it is even possible to calculate the object’s mass if the clockwise and counterclockwise torques can be calculated. For example, if a uniform meter stick is at equilibrium and has a 220g suspended at 5.0cm, 120g at 90.0cm, and the support stand at 40cm, it can be concluded that ((.4m–.05m)*.2kg*9.8m/s2)=((.9m–.4m)*.12kg*9.8m/s2) + ((.5m-.4m)*MASS*9.8m/s2). Solving for mass, we get that the MASS of the meter stick is 0.1kg.
Conclusion: In conclusion, this lab shows how torque can be balanced by manipulating the position of a force to or away from the axis. As predicted in the hypothesis, moving a force closer to the axis decreases torque and moving a force further from the axis increases torque. Table 1.0 shows an example of this as a small mass of 0.097kg can be balanced with a large mass of 0.201kg simply by placing the small mass further from the fulcrum and the large mass closer to the fulcrum. When the clockwise and counterclockwise torques are equal in magnitude, but opposite in direction, a net torque of zero will result in a balanced meter stick. This concept of rotational equilibrium applies to all systems of rigid bodies that rotate around a fixed axis.
Mr. Kellogg
AP Physics - Pd. 7
6 April 2012
Performed On: 4-5 April 2012
Lab Partners: Sir Sebastian Alvarez, Esq., Lt. Lue Vang
Torque and Rotational Equilibrium
Purpose:
This lab will study torque, specifically balancing two levels of torque on two sides of a pivot point to achieve rotational equilibrium. The lab will study the effects of having variable masses at variable distances from the pivot point.
Background:
The lever arm will be the distance of the placed mass away from the pivot point.
The percent difference between two of the same variables of the same units is the difference between the two values divided by the average of the two values.
Hypothesis:
Since rotation is caused by a net torque, balancing the torques on either side of the fulcrum should result in a stable, balanced meter stick. Since torque is r⊥F, moving the masses further from the fulcrum should increase the torque on that side of fulcrum. As it is, a light weight can be balanced with a large weight by placing the light weight closer to the edge of its side and the large weight closer to the balance point.
Apparatus:
Procedure:
Case I: Two Known Masses:
First, balance a meter stick on the fulcrum. Hang a 100g mass at 15.0cm on the meter stick and hang a 200g mass on the other side so that the meter stick is balanced on the fulcrum. Record the data and calculate torque.
Case II: Three Known Masses:
First, balance a meter stick on the fulcrum. Hang a 100g mass at 30.0cm on the meter stick. Then, hang 200g at 70.0cm. Finally hang a 50g mass in a position that will balance the meter stick on its fulcrum. Record the data and calculate torque.
Case III: Unknown Masses:
First, balance a meter stick on the fulcrum. Place one of the unknown masses at 10.0cm and then suspend a 200g mass on the other side so that the meter stick is balanced. Calculate a prediction for the position of the 200g and then test it. Measure and record data.
Case IV: Meter Stick with One Mass:
Suspend 100g at 0.0cm of the meter stick. Slide the fulcrum position on the meter stick so that the meter stick is balanced. Record the data and calculate toqrues.
Data:
Case I: Two Known Masses:
i.e.) 0.502 – 0.150 = 0.352 m
weight = mass * gravity = mg = kg*m/s2 = N
i.e.) 0.097 * 9.8 = 0.951 N
torque = lever arm * weight = rF = m*N
i.e.) -0.352 * 0.951 = -0.335
Case II: Three Known Masses:
i.e.) 0.192 + 0.202 = .394 m*N
clockwise torque = τ2 = τcw
i.e.) -0.390 = -0.390 m*N
percent difference = (difference of magnitude/average)*100
i.e.) 0.004 * 0.39 = 0.99%
A General Rule:
-( -0.098*1.970) + ((0.502-x) * 0.490)= (0.302 * 0.951)
x = .694 m
*x = calculated mass position
i.e.) |(0.694 – 0.690)/.690|*100 = 0.58%
Case III: Unknown Masses:
*Mass B had .04kg added to be able to fit it onto the meter stick
= ((lever arm of moving mass * weight of moving mass)/(lever arm of stationary mass*9.8)) - mass of hanger
= ((rm * wm)/(rs * g)) - mh
= (-(rm * 1.97)/(0.402 * .98)) – 0.017
i.e.) (-(-0.467 * 1.97)/(0.402 * .98)) – 0.017 = .216
Case IV: Meter Stick with One Mass:
Analysis:
According to the data, the torque from a large force can be reduced by moving the force closer to the axis. In vice versa, a small force can produce a larger torque by applying that force farther away from the axis.
Note though, that the meter stick and hanging masses are suspended above the table; thus, there is an upward force equal to all the downward force from gravity in order for the objects to remain elevated. These “support forces” are not included in torque calculations because they cancel out and have a net force of zero, thus no net torque.
All in all, if an object is balanced at rotational equilibrium, it is even possible to calculate the object’s mass if the clockwise and counterclockwise torques can be calculated. For example, if a uniform meter stick is at equilibrium and has a 220g suspended at 5.0cm, 120g at 90.0cm, and the support stand at 40cm, it can be concluded that ((.4m–.05m)*.2kg*9.8m/s2)=((.9m–.4m)*.12kg*9.8m/s2) + ((.5m-.4m)*MASS*9.8m/s2). Solving for mass, we get that the MASS of the meter stick is 0.1kg.
Conclusion:
In conclusion, this lab shows how torque can be balanced by manipulating the position of a force to or away from the axis. As predicted in the hypothesis, moving a force closer to the axis decreases torque and moving a force further from the axis increases torque. Table 1.0 shows an example of this as a small mass of 0.097kg can be balanced with a large mass of 0.201kg simply by placing the small mass further from the fulcrum and the large mass closer to the fulcrum. When the clockwise and counterclockwise torques are equal in magnitude, but opposite in direction, a net torque of zero will result in a balanced meter stick. This concept of rotational equilibrium applies to all systems of rigid bodies that rotate around a fixed axis.